Intuition behind $1-\gamma$ and $\frac{1}{1-\gamma}$ for calculating discounted future state distribution and discounted reward
Question 1
The taylor expansion of $\frac{1}{1-\gamma}$ at $\gamma= 0$ is as follows
$$\frac{1}{1-\gamma} = 1 + \gamma + \gamma^2 + \dots$$
When you multiply by $1-\gamma$ you get
$$ 1 = (1-\gamma)(1 + \gamma + \gamma^2 + \dots)$$
Which can be equivalently written as
$$1 = (1-\gamma)\sum_\limits{i=0}^{\infty}\gamma^i$$
Hence we can see that by multiplying the coefficients by $(1-\gamma)$ we get a weighted sum of transition probabilities

Question 2
Multiplying by $(1-\lambda)$ is to cancel out the normalisation term which was included in the definition of the discounted distribution of the states. Partially expanding out your expressions for the discounted reward you get this
$$ J(\pi) = \frac{1}{1-\gamma}\sum_\limits{s}E_
{a\sim\pi,s'\sim P}[R(s,a,s')]\cdot\left((1-\gamma)\sum_\limits{t=0}^{\infty}\gamma^tP(s_t=s|\pi)\right)$$
The discounted return form is usually denoted
$$J(\pi) = E[\sum\gamma^iR_i(s,a,s')]$$
Where you notice the discounting terms aren't normalised which motivates cancelling it out